Question

find the area of shaded portion
​

Answer 1

Answer:

$1\3$

first you should count all the parts and the see that how many parts are shaded and when you count then so you should write .

Answer 2

Question :-

Find the Area of the Shaded Region.

To Find :-

The Area of the Shaded Region .

Given :-

• AO = 12 units

• OC = 7 units

• AB = 15 units

• CB = 14 units

We Know :-

Area of a Triangle :-

$\underline{\boxed{\bf{A = \dfrac{1}{2} \times Base \times Height}}}$

Area of Scalene Triangle :-

$\underline{\boxed{\bf{A = \sqrt{s(s - a)(s - b)(s - c)}}}}$

Where :-

• a = Side of the Triangle
• b = Side of the Triangle
• c = Side of the Triangle
• s = Semi-perimeter

Semi-Perimeter :-

$\boxed{\bf{s = \dfrac{a + b + c}{2}}}$.

Pythagoras theorem :-

$\boxed{\bf{h^{2} = b^{2} + p^{2}}}$

Where :-

• h = Hypotenuse
• b = Base
• p = Height

Concept :-

A/c :-

Area of ∆ ABC - Area of ∆ ABO = Area of Shaded Region.

But first , we have to find the length of AC , by using the Pythagoras theorem .

And by finding the Area of ∆ABC and the Area of ∆ABO , then by. subtracting them , we will get the required value.

Solution :-

To Find the Length of AC :-

Given :-

• Height = 12 units
• Base = 7 units

Let the hypotenuse be h units.

Using the Pythagoras theorem and substituting the values in it , we get :-

$:\implies \bf{h^{2} = b^{2} + p^{2}} \\ \\ \\ :\implies \bf{h^{2} = 7^{2} + 12^{2}} \\ \\ \\ :\implies \bf{h = \sqrt{7^{2} + 12^{2}}} \\ \\ \\ :\implies \bf{h = \sqrt{49 + 144}} \\ \\ \\ :\implies \bf{h = \sqrt{7^{2} + 13^{2}}} \\ \\ \\ :\implies \bf{h = \sqrt{193}} \\ \\ \\ :\implies \bf{h = 13.9(approx.)} \\ \\ \\ \therefore \purple{\bf{h = 13.9 units}}$.

Hence, the Length of hypotenuse or AC is 13.9 units.

Area of Triangle ABO :-

• Height = 12 units.

• Base = 7 units

Using the formula and substituting the values in it , we get :-

$:\implies \bf{A = \dfrac{1}{2} \times Base \times Height} \\ \\ \\ :\implies \bf{A = \dfrac{1}{2} \times 7 \times 12} \\ \\ \\ :\implies \bf{A = 7 \times 6} \\ \\ \\ :\implies \bf{A = 42 units} \\ \\ \\ \therefore \purple{\bf{A = 42 units²}}$

Hence,the area of the triangle ∆ABO is 42 units².

Area of the triangle ABC :-

• AC = 13.9 units

• AB = 15 units

• CB = 14 units

Semi-Perimeter :-

Using the formula and substituting the values in it ,we get :-

$:\implies \bf{s = \dfrac{a + b + c}{2}} \\ \\ \\ :\implies \bf{s = \dfrac{13.9 + 14 + 15}{2}} \\ \\ \\ :\implies \bf{s = \dfrac{13.9 + 14 + 1}{2}} \\ \\ \\ :\implies \bf{s = \dfrac{42.9}{2}} \\ \\ \\ :\implies \bf{s = 21.5 units} \\ \\ \\ \therefore \purple{\bf{s = 21.5 units}^{2}}$

Thus , The Semi-Perimeter = 21.5 units.

Now ,

we Know that ;

• a = 13.9 units

• b = 15 units

• c = 14 units

• s = 21.5 units

Using the formula and substituting the values in it , we get :-

$:\implies \bf{A = \sqrt{s(s - a)(s - b)(s - c)}} \\ \\ \\ :\implies \bf{A = \sqrt{21.5(21.5 - 13.9)(21.5 - 14)(21.5 - 15)}} \\ \\ \\ :\implies \bf{A = \sqrt{21.5 \times 7.6 \times 7.5 \times 6.5}} \\ \\ \\ :\implies \bf{A = \sqrt{7965.8}} \\ \\ \\ :\implies \bf{A = 89.25} \\ \\ \\ \therefore \purple{\bf{A = 89.25 units^{2}}}$

Hence, the Area of the triangle ∆ABC is 89.25 units².

Area of the Shaded Region :-

==> Area of ∆ABC - Area of Triangle ∆ABO.

==> 89.24 - 42

==> 47.25 units ²

hence, the area if thr shaded region is 47.25 unit².